Category: ICSE Class 10

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E

Prove the following Identities :
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
sin⁴A – cos⁴ A = 2 sin²A-1 
Solution:
L.H.S. = sin⁴ A – cos⁴A = (sin²A)²-(cos²A)²
= (sin²A + cos²A) (sin²A – cos²A)     [(a² – b² = (a + b) (a – b)]
= 1 (sin² A – cos²A) [∵ sin²A + cos²A = 1]
= sin² A – (1- sin²A) (∵ cos²A = 1 – sin²A)
= sin² A – 1 + sin² A
= 2 sin²A-1 = R.H.S.

Question 6.
(1 – tan A)² + (1 + tanA)² = 2sec²A
Solution:
LHS = (1 -tanA)² + (1 +tanA)²
= 1 + tan² A- 2 tan A + 1 + tan² A + 2 tanA
= 2 + 2 tan² A = 2 (1+tan²A)
= 2 sec²A (∵ l+tan²A=sec²A)
= R.H.S.

Question 7.
Cosec⁴ A – cosec² A = cot⁴ A + cot² A
Solution:
L.H.S. = cosec⁴ A -cosec² A
= (cosec²A)² – cosec²A
= (1 + cot²A)² – (1 + cot²A)
= 1 + cot⁴ A + 2 cot²A – 1- cot²A
= cot⁴ A + cot² A = R.H.S.

Question 8.
sec A (1-sin A) (sec A + tan A) = 1
Solution:

Question 9.
cosec A (1 + cos A) (cosec A – cot A) = 1
Solution:

Question 10.
sec² A + cosec²A = sec² A cosec² A
Solution:

Question 11.

Solution:

Question 12.
tan²A – sin²A = tan² A. sin² A
Solution:

Question 13.
cot² A – cos² A = cos² A. cot² A
Solution:

Question 14.
(cosecA + sinA) (cosec A – sinA) = cot² A + cos²A
Solution:
L.H.S. = (cosec A + sin A) (cosec A – sin A)
= (cosec²A – sin² A) [∵ (a + b) (a – b) = a²– b²]
= 1 + cot² A – sin² A = cot² A + 1 – sin²A
= cot² A + cos² A (∵ 1-sin²A = cos² A)
= R.H.S.

Question 15.
(sec A – cosA) (sec A + cosA) = sin² A + tan²A
Solution:
L.H.S. = (sec A-cos A) (sec A + cos A)
= sec² A – cos² A
= 1 + tan²A-cos² A
= 1-cos² A + tan² A
= sin² A + tan² A  (∵ 1- cos²A=sin²A)
= R.H.S.

Question 16.
(cos A + sin A)² + (cos A – sin A)² = 2
Solution:
LHS = (cos A + sin A)² + (cos A – sin A)²
= cos² A + sin² A + 2 cos A sin A + cos² A + sin² A – 2 cos A sin A
= 2 sin² A + 2 cos² A
= 2 (sin²A+cos²A)
= 2 x 1=2 = R.H.S. (∵ sin²A + cos² A = 1)

Question 17.
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.
(sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Solution:
L.H.S. = (sin A + cosecA)² + (cosA+ secA)²
= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
= sin² A+cosec² A+2 sin A x \(\frac { 1 }{ sinA }\) + cos² A+sec²A + 2cosA x \(\frac { 1 }{ cosA }\)
= sin²A + cos² A + cosec² A + sec²A+ 2 + 2   (∵ sin² A + cos²A= 1)
= 1 +cosec²A + sec²A + 4
= (1 + cot² A) + (1 + tan² A) + 5 [∵ cosec²A = 1 + cot² A and sec² A = 1 + tan²A]
= 1 + cot² A + 1 + tan² A + 5
= 7 + tan²A + cot²A = R.H.S.

Question 22.
sec²A. cosec²A = tan²A + cot²A + 2
Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.

Solution:

Question 36.

Solution:

Question 37.

Solution:

Question 38.
(1 +cot A-cosec A) (1 + tan A + sec A) = 2
Solution:

Question 39.

Solution:

Question 40.

Solution:

Question 41.

Solution:

Question 42.

Solution:

Question 43.

Solution:

Question 44.

Solution:

Question 45.

Solution:

Question 46.

Solution:

Question 47.

Solution:

Question 48.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
Find x and y, if:

Solution:

Question 2.
Find x and y, if :

Solution:

Question 3.

Solution:

Question 4.

(i) the order of the matrix X
(ii) the matrix X.
Solution:

Question 5.
Evaluate

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

(i) A (BA)
(ii) (AB) A
Solution:

Question 9.
Find x and y, if

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:
(i) Order of matrix A is 2 x 2
Order of matrix B is 2 x 1
Order of matrix X is 2 x 1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

Question 1.
Given a triangle ABC in which A = (4, -4), B (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.
Solution:
B (0, 5), C (5, 10) and BP : PC = 3 : 2 Co-ordinates of P will be

Question 2.
A (20, 0) and B (10, – 20) are two fixed points, find the co-ordinates of the point P in AB such that 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Solution:
(i) A (20, 0), B (10, – 20)

Question 3.
A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Show that: PQ = \(\frac { 3 }{ 8 }\) BC.
Solution:

Question 4.
Find the co-ordinates of points of trisection of the line segment joining the points (6, -9) and the origin.
Solution:
Points are A (6, -9) and O (0,0) let P and Q are points, which trisect AO

Question 5.
A line segment joining A (-1, \(\frac { 5 }{ 3 }\)) and B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects the y-axis.
(i) Calculate the value of ‘a’.
(ii) Calculate the co-ordinates of ‘P’. (1994)
Solution:

Question 6.
In what ratio is the line joining A (0, 3) and B (4, -1), divided by the x-axis ? Write the co-ordinates of the point where AB intersects the x-axis. [1993]
Solution:
Let the ratio be m₁ : m₂ when the x-axis intersects the line AB at P.

Question 7.
The mid point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B. (1996)

Solution:
Let co-ordinates of A (x, 0) and B (0, y) and C (4, -3) the mid point of AB.

Question 8.
AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7). Find
(i) the length of radius AC
(ii) the coordinates of B.
Solution:

Question 9.
Find the co-ordinates of the centroid of a triangle ABC whose vertices are A (- 1, 3), B (1, – 1) and C (5, 1)
Solution:

Question 10.
The mid-point of the line segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
Solution:
Let A and B are two points and P is its mid point then A is (4a, 2b -3), B(-4, 2b) and P is (2, -2a)

Question 11.
The mid point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
Solution:
The midpoint of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1)

Question 12.
(i) Write down the co-ordinates of the point P that divides the line joining A (-4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB ? [ICSE 1995]
Solution:
Point P, divides a line segment giving the points A (-4, 1) and B (17, 10) is the ratio 1 : 2.

Question 13.
Prove that the points A(-5, 4); B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D. So that ABCD is a square. [1992]
Solution:
In ABC, the co-ordinates of A, B and C are (-5, 4), B(-1, -2) and C (5, 2) respectively.

ABC is also a right-angled triangle.
Hence ABC is an isosceles right angled triangle,
Let D be the fourth vertex of square ABCD and co-ordinates of D be (x,y)
Since the diagonals of a square bisect each other and let O be the point of intersection of AC and BD.
O is mid-point of AC as well as BD.

Question 14.
M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1). Find the co-ordinates of point M. Further, if R (2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q.
Solution:
Two points are given A (-3, 7) and B (9, -1)
M is the mid-point of line joining AB.
Co-ordinates of M wll be

Question 15.
Calculate the ratio in which the line joining A (-4, 2) and B (3, 6) is divided by point P (x, 3). Also find
(i) x
(ii) Length of AP. (2014)
Solution:
Let ratio = k : 1

Question 16.
Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3, 7).
Solution:
Let the given line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3,7) in the ratio k : 1 at a point (x, y) on it.

Question 17.
If the abscissa of a point P is 2. Find the ratio in which this point divides the line segment joining the points (-4, 3) and (6, 3). Also, find the co-ordinate of point P.
Solution:
Abscissa of a point P is 2
Let co-ordinates of point P be (2, y)
Let point P (2, y) divides the line segment joining the points (-4, 3) and (6, 3) in the ratio k : 1

Question 18.
The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k, Also, find the co-ordinates of point Q.
Solution:
A line joining the points (2, 1) and (5, -8) is trisector at P and Q.

Question 19.
M is the mid-point of the line segment joining the points A (0, 4) and B (6, 0). M also divides the line segment OP in the ratio 1 : 3. Find:
(i) co-ordinates of M
(ii) co-ordinates of P
(iii) length of BP

Solution:
M is mid point of the line segment joining the points A (0, 4) and B (6, 0)
M divides the line segment OP in the ratio 1 : 3

Question 20.
Find the image of the point A (5, -3) under reflection in the point P (-1, 3).
Solution:
Image of the point A (5, -3) under reflection in the point P (-1, 3)
Let B (x, y) be the point of reflection of A (5, -3) under P(-1, 3)

Question 21.
A (-4, 2), B (0, 2) and C (-2, -4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of PQR is the same as the centroid of ABC.
Solution:
A (-4, 2), B (0, 2) and C (-2, -4) are the vertices of ABC.
P, Q and R are the mid-points of the sides BC, CA and AB respectively.
G is the centroid of medians AP, BQ and CR.
Co-ordinates of G are

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
Evaluate if possible :

If not possible, give a reason.
Solution:

(iv) It is not possible, because number of columns of the first matrix is not equal to number of rows of the second matrix.

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Find x and y, if:

Solution:

Question 5.

Solution:

Question 6.

(i) AB
(ii) BA
(iii) A²
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:


Comparing the elements, we get:
-2b =-2
b = 1
a = 2

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Simplify : A² + BC.
Solution:

Question 16.
Solve for x and y :

Solution:

Question 17.
In each case given below, find :
(a) the order of matrix M.
(b) the matrix M.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.
If A and B are any two 2 x 2 matrices such that AB = BA = B and B is not a zero matrix, What can you say about the matrix A?
Solution:
AB = BA = B
But it is possible, when A = 0 or B = 0
But B is not a zero matrix (given)
A is a zero matrix or A is an identity matrix

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.
State, with reason, whether the following are true or false. A, B and C are matrices of order 2 x 2.
(i) A + B = B + A
(ii) A – B = B – A
(iii) (B . C). A = B . (C . A)
(iv) (A + B) . C = A . C + B . C
(v) A . (B – C) = A . B – A . C
(vi) (A – B) . C = A . C – B . C
(vii) A² – B² = (A + B) (A – B)
(viii) (A – B)² = A² – 2 A . B + B²
Solution:
(i) True : Because addition of matrices is commutative.
(ii) False : Subtraction of matrices is not commutative.
(iii) True : Multiplication of matrices is associative.
(iv) True: Multiplication of matrices is distributive over addition.
(v) True : As given above in (iv)
(vi) True : As given above in (iv)
(vii) False : Laws of algebra for factorization and expansion are not applicable to matrices.
(viii) False, As given above in (vii)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

Question 1.
Find the mid-point of the line segment joining the points:
(i) (-6, 7) and (3, 5)
(ii) (5, -3), (-1, 7)
Solution:
Let P (x, y) be the mid-point in each case

Question 2.
Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.
Solution:
Co-ordinates of A (3, 5), B (x, y) and mid-point M (2, 3)

Question 3.
A (5, 3), B (-1, 1) and C (7, -3) are the vertices of ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = \(\frac { 1 }{ 2 }\) BC.
Solution:

Question 4.
Given M is the mid-point of AB, find the co-ordinates of:
(i) A; if M = (1, 7) and B = (-5, 10),
(ii) B; if A = (3, -1) and M (-1, 3).
Solution:
M is the mid-point of AB.
(i) Let A = (x, y), M = (1, 7) and B = (-5, 10)

Question 5.
P (-3, 2) is the mid-point of line segment AB as shown in the figure. Find the co-ordinates of points A and B.

Solution:

Point A is on y-axis
its abscissa is zero and point B is on x-axis
its ordinate is zero.
Now, let co-ordinates of A are (0, y) and ofB are (x, 0) and P (-3, 2) is the mid-point

Question 6.
In the given figure, P (4, 2) is the mid point of line segment AB. Find the co-ordinates of A and B.

Solution:

Points A and B are on x-axis and y-axis respectively
Ordinate of A is zero and abscissa of B is zero.
Let co-ordinates of A be (x, 0) and B (0, y)
and P (4, 2) is the mid-point

Question 7.
(-5, 2), (3, -6) and (7, 4) arc the vertices of a triangle. Find the length of its median through the vertex (3, -6) and (7, 4).
Solution:

Let A (-5, 2), B (3, -6) and C (7, 4) are the vertices of a ABC
Let L,M and N are the mid-points of sides BC, CA and AB respectively of ABC.
L is the mid-point of BC.
Co-ordinates of L will be

Question 8.
Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Solution:

Question 9.
One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).
Solution:

Question 10.
A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
Solution:
Co-ordinates of A = (2, 5), B = (1, 0), C = (-4, 3) and D = ( 3, 8)

Let the mid-point of AC is P (x1, y1) Co-ordinates of mid-point of AC will be

Co-ordinates of mid-points AC and BD are the same..
The quadrilateral is a parallelogram.

Question 11.
P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of its diagonals. Find the co-ordinates of R and S.
Solution:
In the parallelogram PQRS and qo-ordinates of P are (4, 2) and of Q are (-1, 5).
The diagonals of || gm AC and BD intersect each other at O (-3, 2)

Question 12.
A (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the co-ordinates of vertex C.
Solution:

Vertices of a parallelogram ABCD are A (-1, 0), B (1, 3) and D(3, 5)
Let co-ordinates of C be (x, y)
Let the diagonals AC and BD bisect each other at O. Then O is the mid-point of AC as well as of BD.
Co-ordinates of O, the mid-point of BD will be

Question 13.
The points (2, -1), (-1, 4) and (-2, 2) are the mid-points of the sides of a triangle. Find its vertices.
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of a ABC respectively.


Co-ordinates of A are (-5, 7), of B are (1, -3) and of C are (3, 1)

Question 14.
Points A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e. lie on the same straight line) such that AB = BC. Calculate the values of x and y.
Solution:

Question 15.
Points P (a, -4), Q (-2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of ‘a’ and ‘b’:
Solution:

Question 16.
Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, -2), B = (0, 1) and C = (-1, 4).
Solution:

Question 17.
The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
Solution:

Question 18.
A (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
Evaluate

Solution:

Question 2.
Find x and y if :

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If I is the unit matrix of order 2 x 2; find the matrix M, such that:

Solution:

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Question 1.
Find AD.

Solution:

Question 2.
In the following diagram.
AB is a floor-board. PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.

Solution:

Question 3.
Calculate BC.

Solution:

Question 4.
Calculate AB .

Solution:

Question 5.
The radius of a circle is given as 15cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, Calculate :
(i) the length of AB;
(ii) the distance of AB from the centre C.
Solution:
Chord AD substends an angle of 131° at the centre. Join CA, CB and draw CD ⊥ AB which bisects AB at D.
(In ∆CAB)
∵ CA = CB (radii of the same circle)
∴ ∠CAB = ∠CBA
But ∠CAB + ∠CBA = 180°- 131° = 49°

Question 6.
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \(\frac { 5 }{ 12 }\). On walking 192 metres towards the tower; the tangent of the angle is found to be \(\frac { 3 }{ 4 }\). Find the height of the tower.
Solution:

Question 7.
A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is a and that of the top of flagstaff is β. Prove that the height of the tower is :

Solution:

Question 8.
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m.
The man’s eye is 2 m above the ground. He observes the angle of elevation of C. The top of the pole, as x°, where tan x° = \(\frac { 2 }{ 5 }\) . calculate:

(i) the distance AB in m;
(ii) the angle of elevation of the top of the pole
when he is standing 15 m from the pole. Give your answer to the nearest degree. [1999]
Solution:

Question 9.
The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same striaght line with it are complementary.Prove that height of the tower is \( \sqrt{ab} \) metre.
Solution:

Question 10.
From a window A. 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x = \(\frac { 5 }{ 2 }\) and the angle of depression of the foot D of the tower is y°, where tany° = \(\frac { 1 }{ 4 }\).(See the figure given below). Calculate the height CD of the tower in metres. [2000]

Solution:
Let CD be the height of the tower and height of window A from the ground = 10m
In right ∆AEC,

Question 11.
A vertical tower is 20 m high, A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? [2001]
Solution:

Question 12.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate :
(i) the width of the river and
(ii) the height of the tree.
Solution:

Let TR be the tree of height x m and y be the width AR of the river,
then ∠B = 30° and A = ∠60° , AB = 50 m.
Now in right ∆ATR,

Question 13.
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is 60° and the angle of elevation of the top of the pole as seen from the foot of the tower is 30°. Find
(i) the height of the tower.
(ii) the horizontal distance between the pole and the tower.
Solution:
Let PQ is the pole and TS is the tower. PQ = 20 m.
Let TS = h and QS = x Angles of elevation from Q to T A T is 60° and from S to P is 30°.
In the ∆PQS

Question 14.
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°.
Find : (i) the height of the tower, if the height of the pole is 20m;
(ii) the height of the pole, if the height of the tower is 75 m.
Solution:
Let PQ is the pole and TW is the tower
Angle of elevation from T to P is 60° and angle of depression from P to W is 30°
∴ ∠PWQ = 30° = ∠RPW ( ∠ Altanate angles)
(i) In first case when height of pole OQ = 20m, Then in right ∆ PQW

Question 15.
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.
Solution:
Let AQ is the sea-level
P is a point 36 m above sea-level
∴ PQ = 36
Let B be the bird and R is its reflection in the water and angle of elevation of the bird B at P is 30° and angle of depression of the reflection of the bird at R is 60°

Question 16.

Solution:

Question 17.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of the light house in horizontal line with its base , are 30° and 40° respectively . Find the distance between the two ships.Give your answer correct to the nearest meter. [2012]
Solution:

Question 18.
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :

(i) the horizontal distance between AB and CD.
(it) the height of the lamp post.
Solution:

Question 19.
An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:

Question 20.
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24° respectively. Find the height of the two towers. Give your answer correct to 3 significant figures. (2015)

Solution:
AB and CD are two towers which are 120 m apart
i.e. BD= 120m
Angles of elevation of the top and angle of depression of bottom of the first tower observed from the top of second tower is 30° and 24°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E

Question 1.
A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’. Calculate:
(i) the length of AB, if A’ B’ = 6 cm.
(ii) the length of C’ A’ if CA = 4 cm.
Solution:
Scale factor (k) = 2.5
∆ABC is enlarged to ∆A’B’C’
(i) A’B’ = 6 cm

Question 2.
A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:
(i) the length of M’ N’, if MN = 8 cm.
(ii) the length of LM, if L’ M’ = 5.4 cm.
Solution:
∆LMN has been reduced by the scale factor

Question 3.
A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find:
(i) A’ B’, if AB = 4 cm.
(ii) BC, if B’ C’ = 15 cm.
(iii) OA, if OA’= 6 cm.
(iv) OC’, if OC = 21 cm.
Also, state the value of:
(a) \(\frac { OB’ }{ OB }\)
(b) \(\frac { C’A’ }{ CA }\)
Solution:
∆ABC is enlarged to ∆A’B’C’ about the point O as its centre of enlargement.
Scale factor = 3 = \(\frac { 3 }{ 1 }\)

Question 4.
A model of an aeroplane is made to a scale of 1 : 400. Calculate:
(i) the length, in cm, of the model; if the length of the aeroplane is 40 m.
(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.
Solution:
Model of an aeroplane to the actual = 1 : 400

Question 5.
The dimensions of the model of a multistorey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30; find the actual dimensions of the building.
Solution:
Dimensions of a model of multistorey building = 1.2 m x 75 cm x 2 m

Question 6.
On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.
Calculate:
(i) the actual lengths of AB and BC in km.
(ii) the area of the plot in sq. km.
Solution:
Scale of map drawn of a triangular plot = 1 : 2,50,000
Measurement of plot AB = 3 cm, BC = 4 cm
and ∠ABC = 90°

Question 7.
A model of a ship of made to a scale 1 : 300
(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.
(ii) The area of the deck ship is 180,000 m². Calculate the area of the deck of the model.
(iii) The volume of the model in 6.5 m3. Calculate the volume of the ship. (2016)
Solution:

Question 8.
An aeroplane is 30 in long and its model is 15 cm long. If the total outer surface area of the model is 150 cm², find the cost of painting the outer surface of the aeroplane at the rate of ₹ 120 per sq.m. Given that 50 sq. m of the surface of the aeroplane is left for windows.
Solution:
Length of aeroplane = 30 m = 3000 cm
and length of its model = 15 cm
Surface area of model = 150 cm²
Scale factor (k) = \(\frac { 3000 }{ 15 }\) = \(\frac { 200 }{ 1}\)
Area of plane = k² x area of model = (200)² x 150 cm² = 40000 x 150 cm²
\(\frac { 40000 x 150 }{ 10000 }\) = 600 m² (1 m² = 10000 cm²)
Shape left for windows = 50 sq. m
Balance area = 600 – 50 = 550 sq. m
Race of painting the outer surface = ₹ 120 per sq.m
Total cost = ₹ 550 x 120 = ₹ 66000

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15D are helpful to complete your math homework.

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Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E

Question 1.
Prove the following identities :

Solution:

Solution:

Solution:

Solution:

Solution:


Solution:

Solution:

Solution:


Solution:

Solution:

Solution:

Solution:


Solution:

Solution:

Solution:


Solution:

Solution:

Question 2.
If sin A + cos A = p and sec A + cosec A = q then prove that: q(p² – 1) 2p
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If tan A=n tan B and sin A=m sin B, prove that:

Solution:

Question 6.
(i) If 2 sin A-1 = 0, show that:
sin 3 A = 3 sin A – 4 sin³ A.             [2001]
(ii) If 4cos² A-3 = 0, show that:
cos 3A = 4 cos³ A – 3 cos A
Solution:

Question 7.
Evaluate:

Solution:

Question 8.
Prove that:

Solution:

Question 9.
If A and B are complementary angles, prove that:
(i) cot B + cos B sec A cos B (1 + sin B)
(ii) cot A cot B – sin A cos B – cos sin B = 0
(iii) cosec² A + cosec² B = cosec² A cosec² B

Solution:

Question 10.
Prove that:


Solution:

Question 11.
If 4 cos² A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that : 
(i) sin 3A = 3 sinA – 4 sin³A
(ii) cos 3A = 4 cos³ A – 3 cos A
Solution:

Question 12.
Find A, if 0° ≤ A ≤ 90° and :
(i) 2 cos² A – 1 = 0
(ii) sin 3A – 1 = 0
(iii) 4 sin² A – 3 = 0
(iv) cos² A – cos A = 0
(v) 2cos² A + cos A – 1 = 0
Solution:

Question 13.
If 0° < A < 90° ; find A, if :

Solution:

Question 14.
Prove that : (cosec A – sin A) (sec A – cos A) sec² A = tan A. (2011)
Solution:

Question 15.
Prove the identity : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. (2014)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9D

Question 1.
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 x 2 and 2 x 3 respectively; then their sum A + B is possible,
(ii) The matrices A 2 x3, and B 2 x 3, are conformable for subtraction,
(iii) Transpose of a 2 x 1 matrix is a 2 x 1 matrix,
(iv) Transpose of a square matrix is a square matrix,
(v) A column matrix has many columns and only one row.
Solution:
(i) False, because the orders of both the matrices are not same,
(ii) True,
(iii) False, because transpose of a 2 x 1 matrix is a 1 x 2 matrix,
(iv) True,
(v) False, because it has only one column and may have many rows.

Question 2.

Solution:
Comparing the elements in order, we get that :
x = 3
y + 2 = 1 ⇒ y = 1 – 2 = -1
z – 1 = 2 ⇒ z = 2 + 1 = 3
Hence x = 3, y = -1 , z = 3

Question 3.
Solve for a, b and c; if :

Solution:
(i) Comparing the elements in order, we get
-4 = b + 4 ⇒ – b = 4 + 4 = 8 ⇒ b = – 8
a + 5 = 2 ⇒ a = 2 – 5= -3
2 = c – 1 ⇒ c = 2 + 1 = 3
Hence a = -3, b = -8, c = 3
(ii) Comparing the elements in order, we get
a = 3
a – b = – 1 ⇒ 3 – b = -1 ⇒ -b = -1 – 3 ⇒ -b = -4 ⇒ b = 4
b + c = 2 ⇒ 4 + c = 2 ⇒ c = 2 – 4 = -2
Hence a = 3, b = 4, c = – 2

Question 4.

Solution:

Question 5.

Solution:

Question 6.
Wherever possible, write each of the following as a single matrix.

Solution:

This is not possible because both the matrices are not of the same order.

Question 7.
Find, x and y from the following equations:

(ii) [-8 x] + [y -2] = [-3 2]
Solution:

Comparing, the elements of two equal martrices:
3 – x = 7 ⇒ x = -7 + 3 = -4
and y + 2 = 2 ⇒ y = 2 – 2 = 0
Hence x = – 4, y = 0
(ii) [-8 x] + [y -2] = [-3 2]
⇒ [-8 + y x – 2] = [-3 2]
Comparing the elements, we get :
-8 + y = -3 ⇒ y = -3 + 8 = 5
x – 2 = 2 ⇒ x = 2 + 2 = 4
Hence x = 4, y = 5

Question 8.

Solution:

Question 9.
Write the additive inverse of matrices A, B and C

Solution:

Question 10.
Given A = [ 2 -3], B = [0 2] and C = [-1 4]; find the matrix X in each of the following :
(i) X + B = C – A
(ii) A – X = B + C
Solution:
A = [2 -3], B = [0 2], C = [-1 4]
(i) X + B = C – A
X + [0 2] = [-1 4] – [2 -3]
⇒ X + [0 2] = [- 1 – 2 4 – (-3)]
X + [0 2] = [-3 4 + 3] = [-3 7]
X = [-3 7] – [0 2] = [-3 – 0 7 – 2] = [-3 5]
(ii) A – X = B + C
⇒ -X = B + C – A
⇒ X = A – B – C = [2 -3] – [0 2] – [-1 4] = [2 – 0 – (-1) -3 – 2 – 4] = [ 2 + 1 – 9] = [3 – 9]

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

Question 1.
Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio, 3 : 5. Find its co-ordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7
Solution:

Question 2.
The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.
Solution:
Point P, divides the line segment A (3, -4) and B(-2, 1) in the ratio of 1 : 3
Let co-ordinates of P be (x, y), then

Question 3.
A line 5x + 3y + 15 = 0 meets y -axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:
P lies on y-axis and let the co-ordinates of P be (0, y)
P lies also on the line 5x + 3y + 15 = 0 it will satisfy it.
5 x 0 + 3y + 15 = 0
⇒ 3y = -15
⇒ y = -5
Co-ordinates of P are (0, -5)
Now, writing the line x – 3y + 4 = 0 is form of y = mx + c
-3y = -x – 4
⇒ 3y = x + 4
⇒ y = \(\frac { 1 }{ 3 }\) x + \(\frac { 4 }{ 3 }\)

Question 4.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other. [2003]
Solution:
Writing, the line kx – 5y + 4 = 0 in form of y = mx + c
⇒ -5y = -kx – 4
⇒ 5y = kx + 4

Question 5.
A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid point of the segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M. (2003)

Solution:


(ii) Let. co-ordinates of A be (x, 0) and of B be (0, y) which lie On the line.
Substituting, the co-ordinates in (i)
x + 0 = 3 ⇒x = 3
Co-ordinates of A are (3, 0)
Again 0 + y = 3 ⇒ y = 3
Co-ordinates of B are (0,3)
(iii) M is the mid-point of AB.
Co-ordinates of M wil be (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

Question 6.
(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:
Co-ordinates of A and C of rhombus ABCD are (1, 5) and (-3, -1)


⇒ 3y – 6 = -2x – 2
⇒ 2x + 3y = 6 – 2
⇒ 2x + 3y = 4

Question 7.
Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:
Three vertices of a square ABCD are A (3, 2), B (6, -2) and C (2, -5)
Let, co-ordinates of fourth vertex D be (x, y)

Question 8.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:
Slope of line x = 3y + 2 or 3y = x – 2 ….(i)

Question 9.
A straight line passes through the point (3, 2) and the portion of this line intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:

Let, the line intersects x-axis at A and y-axis at B.
Let, co-ordinates of A (x, o) and of (o, y)
But (3, 2) is the mid-point of AB.

Question 10.
Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x -2y = 1.
Solution:
7x + 6y = 71 ….(i)
5x – 8y = -23 ….(ii)
Multiply (i) by 4 and (ii) by 3,
28x + 24y = 284
15x – 24y = -69
On adding (i) and (ii), we get:
43x = 215
x = 5
Substituting, the value of x in (i)
7 x 5 + 6y = 71
35 + 6y = 71
⇒ 6y = 71 – 35 = 36
⇒ y = 6
Point of intersection of these lines is (5, 6)
Now slope of line 4x – 2y = 1
⇒ 4x – 1 = 2y
⇒ y = 2x – \(\frac { 1 }{ 2 }\) is 2
Slope of line through the point of intersection and perpendicular to 4x – 2y = 1 is \(\frac { -1 }{ 2 }\)
Equation of the line y – y₁ = m (x – x₁)
⇒ y – 6 = \(\frac { -1 }{ 2 }\) (x – 5)
⇒ 2y – 12 = -x + 5
⇒ x + 2y = 5 + 12 = 17
⇒ x + 2y = 17

Question 11.
Find the equation of the line which is perpendicular to the line \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = 1 at the point where this line meets y-axis.
Solution:

Question 12.
O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of ∆OAB through vertex O.
(ii) the equation of altitude of ∆OAB through vertex B.
Solution:
(i) Let, mid-point of AB be D.

Question 13.
Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does line 3x = y + 1 bisect the line segment joining the two given points ?
Solution:
Slope of the line joining the points (-2, 3) and (4, 1)

Yes, these are perpendicular to each other
Let P be the mid-point of the line joining the points (-2, 3) and (4, 1)
Co-ordinates of P will be

This point (1, 2) satisfies the equaion 3x = y + 1 then, it will bisect the line joining the given point
now, substituting the value of x and y. in 3x = y + 1
⇒ 3 x 1 = 2 + 1
⇒ 3 = 3. which is true.
Yes, the line 3x = y + 1 is the bisector.

Question 14.
Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).
Solution:
Equation of the given line is x cos 30° + y sin 30° = 2
y sin 36° = -x cos 30° + 2

Question 15.
Find the value of k such that the line (k – 2) x + (k + 3) y – 5 = 0 is :
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Writing the given equation in the form of y = mx + c
(k – 2) x + (k + 3) y – 5 = 0 ….(i)
⇒ (k + 3) y = – (k – 2) x + 5

Question 16.
The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7) write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i) meets BC.
Solution:
Vertices of ∆ABC are A (0, 5),B (-1, -2) and C (11, 7)

(ii) Let, the line through A meets BC in P
P is point of intersection of these two lines.
3x – 4y = 5 ……… (i)
4x + 3y = 15 …….. (ii)
On solving (i), (ii) we get
x = 3, y = 1
Co-ordinates of Pare (3, 1)

Question 17.
From the given figure, find :
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC. (2005)

Solution:
(i) From the figure, we see that co-ordinates of A are (2, 3), of B are (-1, 2) of C and (3, 0)
(ii) Slope of line BC is (m)

⇒ x + 2y – 6 – 2 = 0
⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 18.
P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R. (2004)
Solution:
Let (x, y) be the co-ordinates of M, the mid-point of PQ.


⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 19.
A (8, -6), B (-4, 2) and C (0, -10) are the vertjces of a triangle ABC. If P is the mid-point of AR and Q is the mid-point of AC, use co ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.
Solution:
In ∆ABC, co-ordinates of A, B and C are (8, -6), (-4, 2) and C (0, -10) respectively.
P and Q are the mid-points of AB and AC respectively
Co-ordinates of P will be


Slopes of PQ and BC are same.
These are parallel to each other.
Quad. PBCQ is trapezium

Question 20.
A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find :
(i) the co-ordinates of A and B.
(ii) equation of line through P, and perpendicular to AB.
Solution:
Line AB intersects x-axis at A and y-axis at B.
(i) Let co-ordinates of A be (x, 0) and of B be (0, y)
Point P (-4, -2) intersects AB in the ratio 1 : 2

Question 21.
A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from die positive side of y-axis. Find the equation of the line. (1992)
Solution:
Let line intersects x-axis at P (-2, 0) and cuts off an intercept of 3 units at Q.

Question 22.
Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. (2002)
Solution:
x-intercept = 4
Co-ordinates of that point = (4, 0)
The co-ordinates of the given point (2, 3)

Question 23.
The given figure (not drawn to scale) shows two straight lines AB and CD. If’ equation of the line AB is : y = x + 1 and equation of line CD is : y = √3 x – 1. Write down the inclination of lines AB and CD; also, find the angle 6 between AB and CD. (1989)

Solution:
Equation of line AB is y = x + 1
and equation of line CD is y = √3 x – 1
Slope of AB = 1
tanθ = 1
⇒ θ = 45°
Inclination angles of AB = 45°
Slope of CD = tanθ = √3 = tan 60°
⇒ θ = 60°
Inclination angle of CD = 60°
In ΔPQR,
Ext. ∠RQX = ∠RPQ + ∠PRQ (Exterior angles is equal to sum of its interior opposite angles)
⇒ 60° = 45° + θ
⇒ θ = 60° – 45° = 15°
⇒ θ = 15°

Question 24.
Write down the equation of the line whose gradient is \(\frac { 3 }{ 2 }\) and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (1996, 2001)
Solution:
P divides the line segment AB in which A (-2, 6) and B (3, -4) in the ratio 2 : 3
Co-ordinates of P will be

Question 25.
The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Solution:
Let points are A (6, 4) and B (7, -5)

Question 26.
Points A and B have coordinates (7, -3) and (1, 9) respectively. Find
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ of (-2, p) lies on it.
Solution:
Coordinates of A are (7, -3), of B = ( 1, 9)

Question 27.
A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid point of AB. Find the
(i) Coordinates of A and B.
(ii) Slope of line AB.
(iii) equation of line AB.

Solution:
As P (2, -3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)

Question 28.
The equation of a line is 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Given line 3x + 4y -7 = 0


⇒ 3 (y – 4) = 4 (x – 2)
⇒ 3y – 12 = 4x – 8
⇒ 4x – 3y – 8 + 12 = 0
⇒ 4x – 3y + 4 = 0

Question 29.
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find :
(i) co-ordinates of A
(ii) equation of diagonal BD.
Solution:

(i) In || gm ABCD, A (x, y), B (5, 8), C (4, 7) and D (2, -4)
The diagonals of ||gm bisect each other
O is said point of AC and BD
Now if O is mid point of BD then its co-ordinates will be


⇒ y + 4 = 4x – 8
⇒ 4x – y -8 – 4 = 0
⇒ 4x – y – 12 = 0 or 4x – y = 12

Question 30.
Given equation of line L₁ is y = 4.
(i) Write the slope of line L₂ if L₂ is the bisector of angle O.
(ii) Write the co-ordinates of point P.
(iii) Find the equation of L₂.
Solution:
(i) Equation of line L₁ is y = 4
L₂ is the bisector of ∠O

Question 31.
Find:

(i) equation of AB
(ii) equation of CD
Solution:
Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.

Question 32.
Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1
Solution:
x-intercept of the line = -3
and is perpendicular to the line
3x + 5y = 1
5y = 1 – 3x

Question 33.
A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M.
Solution:
A line passing through the two points P (-1, 4) and Q (5, -2) intersects x-axis at point A and y- axis at point B.
M is mid-point of AB

(i) Equation of line AB will be
y – y₁ = m (x – x₁)
⇒ y – 4 = -1 (x + 1)
⇒ y – 4 = – x – 1
⇒ y + x = -1 + 4
⇒ x + y = 3
(ii) The line intersect x-axis at A and OA = 3 units
Co-ordinates of A are (3, 0) and the line intersects y-axis at B and OB = 3 units
Co-ordinates of B are (0, 3)
(iii) M is mid-point of AB
Co-ordinates of M are (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

Question 34.
In the given figure, line AB meets y-axis at point A. Line through C (2, 10) and D intersects line AB at right angle at point P. Find:

(i) equation of line AB.
(ii) equation of line CD.
(iii) co-ordinates of points E and D.
Solution:
In the given figure, AB meets y-axis at point A.
Line through C (2, 10) and D intersects line AB at P at right angle.

Equation of CD
y – 10 = 3 (x – 2)
⇒ y – 10 = 3x – 6
⇒ 3x – y + 10 – 6 = 0
⇒ 3x – y + 4 = 0
(iii) Co-ordinates of D which is on x-axis
3x – y + 4 = 0
3x – 0 + 4 = 0
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = \(\frac { -4 }{ 3 }\)
Co-ordinates of D are (\(\frac { -4 }{ 3 }\) , 0)
E is also on x-axis
x + 3y = 18
Substituting, y = 0, then
x + 0= 18
⇒ x = 18
Co-ordinates of E are (18, 6)

Question 35.
A line through point P (4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Solution:
A line through P (4, 3) meets x-axis at A and the y-axis at B. If BP is double of PA.
Draw BC || x-axis
and PC || y-axis
In ∆PAD and ∆PBC
∠P = ∠P (common)
∠D = ∠C (each 90°)
∆PAD ~ ∆PBC
PB = 2PA ⇒ PA = \(\frac { 1 }{ 2 }\) PB


2y – 6 = 3x – 12
⇒ 3x – 2y – 12 + 6 = 0
⇒ 3x – 2y – 6 = 0
⇒ 3x – 2y = 6

Question 36.
Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Solution:
Equation of given two intersecting lines are 2x – y = 1 and 3x + 2y = -9 Which make an angle of 30°

Question 37.
Find the equation of the line through the points A (-1, 3) and B (0, 2). Hence, show that the points A, B and C (1, 1) are collinear.
Solution:
The given points are A (-1, 3) and B (0, 2) and co-ordinates of a point C are (1, 1)
Now slope of the line joining A and B

Equation of the line y – y₁ = m(x – x₁)
⇒ y – 2 = -1 (x – 0)
⇒ y – 2 = -x
⇒ x + y – 2 = 0
Point C (1, 1) will be on AB if it satisfy
1 + 1 – 2 = 0
⇒ 0 = 0
Point C lies on AB
Hence points A, C and B are collinear.

Question 38.
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2), find:
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)
Solution:
Three vertices of a ||gm ABCD taken an order are A (3, 6), B (5, 10) and C (3, 2)
Join diagonals AC and BD which bisect each other at O.
O is mid-point of AC as well as of BD
Now co-ordinates of O will be

Question 39.
In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.

(i) Write the co-ordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC. (2015)
Solution:
In the given figure,
ABC is a triangle and BC || y-axis
AB and AC intersect the y-axis at P and Q respectively.
(i) Co-ordinates of A are (4,0).
(ii) Length of AB

Question 40.
The slope of a line joining P (6, k) and Q (1 – 3k, 3) is \(\frac { 1 }{ 2 }\). Find :
(i) k
(ii) mid-point of PQ, using the value of ‘A’ found in (i). (2016)
Solution:
(i) Slope of the line joining P(6, k) and Q (1 –

Question 41.
A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB.

Solution:
(i) Since, A lies on the x-axis,
let the coordinates of A be (x, 0).
Since B lies on the y-axis,
let the coordinates of B be (0, y).
Let m = 1 and n = 2.
Using section formula,

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